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I wish to calculate $\oint_{C}\frac{dz}{z(z-1)(z-2)}$ when $C$ is a circle around the origin with radius $1.5$.

I guess that I should somehow apply Cauchy's integral formula here, but $\frac{1}{z},\frac{1}{z-1}$ are not analytical inside of $C$ so I can't define something like $f(z)=\frac{1}{z(z-2)}$ and calculate $\oint_{C}\frac{f(z)dz}{(z-1)}$ by Cauchy's.

Can someone please help me understand how to calculate this integral ?

I am guessing there is some trick so I can use Cauchy's integral formula, but I didn't manage to think of any such tricks.

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Split your circle into two curves, each one surrounding just one singularity (or use the residue theorem). – mrf Jan 10 '13 at 23:32
up vote 3 down vote accepted

$$ \oint_c \frac{1}{z(z-1)(z-2)} dz = 2\pi i \; \left [ \; \text{Res}\left (f(z), 0\right ) + \text{Res}\left ( f(z), 1 \right ) \; \right ] = 2\pi i(1/2 - 1) = -\pi i $$

Or simply, since pole of $z=2 > 1.5 $

$$ \oint_c \frac{1}{z(z-1)(z-2)} dz = \oint_c \left( \frac{1}{2 (-2+z)}-\frac{1}{-1+z}+\frac{1}{2 z} \right ) dz \\ = \oint_c \frac{1}{2z}dz - \oint_c\frac{1}{z-1} = \pi i - 2 \pi i = -\pi i$$

I hope I am not wrong!!

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Thanks! How did you calculate the last two integrals (last line) ? – Belgi Jan 11 '13 at 15:01
@Belgi check out this and this – Santosh Linkha Jan 11 '13 at 22:29
Thanks, you are very kind! By the way: did you use the standard way of finding how to write it with partial fractions ? (or maybe you had a shortcut here ?) – Belgi Jan 12 '13 at 0:55
@Belgi lol ... no, the standard way is to use Residue theorem, since i don't think we will be able to make use of partial fraction every time. I wrote it here since it was possible ... and it's easy to understand it. – Santosh Linkha Jan 12 '13 at 0:58

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